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A parametrization $\phi(u,v)$ of a regular surface is called a 'conformal parametrization if $X_u . X_u$ = $X_v . X_v$ and $X_u . X_v=0$, i.e., $E=G$ and $F=0$. Let $S$ be a regular surface parametrized by a conformal parametrization. Show that if the Gaussian curvature and the mean curvature are identically zero all over $S$, then $S$ is a part of a plane in $\mathbb{R}^3$.

I know that if the gaussian curvature is zero, then $$\frac{LN-M^2}{EG-F^2}=0$$ where $L, M, N$ are the coefficients of the second fundamental form. So, $LN=M^2$. And if the mean curvature is zero too, then $$\frac{LG - 2MF + NE}{2(EG-F^2)}=0$$ So, $LG + NE = 2MF$.

But $F=0$ and $E=G$, so $L = - N$.

Then $-L^2=M^2$, so $L=M=N=0$. But I don't know how to continue.

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We've that $dN_p (X_u) = \lambda X_u$, because the points are all umbilical. But you have just showed that $\langle dN_p (X_u) , X_u \rangle = 0 $. Thus, we have that $dN_p (X_u) = 0$. The same arguing shows that $dN_p (X_v) = 0$. This plainly implies that $N$ is constant, and we've that $S$ is contained in a plane, as desired.

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Another approach:

Since the curvatures are identically zero, the second fundamental form is identically zero. Hence, $(dN)_p = 0$ for all $p \in S$. Now, since $S$ is connected, this implies that $N$ is constant. Hence $S$ is contained in some plane that has $N$ as normal vector.

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